Problem: Simplify; express your answer in exponential form. Assume $t\neq 0, r\neq 0$. $\dfrac{{(t)^{-3}}}{{(t^{-2}r^{-2})^{-4}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${t}$ to the exponent ${-3}$ . Now ${1 \times -3 = -3}$ , so ${(t)^{-3} = t^{-3}}$ In the denominator, we can use the distributive property of exponents. ${(t^{-2}r^{-2})^{-4} = (t^{-2})^{-4}(r^{-2})^{-4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t)^{-3}}}{{(t^{-2}r^{-2})^{-4}}} = \dfrac{{t^{-3}}}{{t^{8}r^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{-3}}}{{t^{8}r^{8}}} = \dfrac{{t^{-3}}}{{t^{8}}} \cdot \dfrac{{1}}{{r^{8}}} = t^{{-3} - {8}} \cdot r^{- {8}} = t^{-11}r^{-8}$.